Determine the molarity and normality of a solution prepared by dissolving 55 mg of sodium bicarbonate (NaHCO3) in 750 mL of water.

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To determine the molarity and normality of a solution prepared by dissolving 55 mg of sodium bicarbonate (NaHCO3) in 750 mL of water, we need to follow a step-by-step process.

Step 1: Convert the mass of sodium bicarbonate (NaHCO3) to moles.
To do this, we need to know the molar mass of NaHCO3, which is calculated by summing the atomic masses of each element present in the compound.

Molar mass of Na = 22.99 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol (There are three oxygen atoms in NaHCO3, so we multiply by 3.)

Now, calculate the molar mass of NaHCO3:
Molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 84.01 g/mol

Next, convert 55 mg to grams:
55 mg = 0.055 g

Now, convert the mass of NaHCO3 to moles using the formula:
moles = mass / molar mass

moles of NaHCO3 = 0.055 g / 84.01 g/mol
= 0.0006549 mol

Step 2: Calculate the molarity of the solution.
Molarity (M) is defined as the number of moles of solute per liter of solution (moles/L).

In this case, we are given that the solution is prepared by dissolving 55 mg of NaHCO3 in 750 mL of water. However, the molarity is usually expressed in terms of liters, so we need to convert 750 mL to liters by dividing by 1000.

Volume of solution (L) = 750 mL / 1000
= 0.75 L

Now, calculate the molarity using the formula:
Molarity = moles of solute / volume of solution

Molarity (M) = 0.0006549 mol / 0.75 L
≈ 0.000873 mol/L

Therefore, the molarity of the solution is approximately 0.000873 mol/L.

Step 3: Calculate the normality of the solution.
Normality (N) is a measure of the number of equivalents of solute per liter of solution. It is used when dealing with acid-base reactions or reactions involving ions with multiple charges.

In the case of sodium bicarbonate (NaHCO3), it has only one ionizable hydrogen atom, which means one mole of NaHCO3 can produce one equivalent of hydrogen ions (H+).

Normality (N) = Molarity (M) × Number of equivalents
(Number of equivalents = 1 for NaHCO3)

Therefore, the normality of the solution is equal to its molarity.

Normality (N) = 0.000873 eq/L

So, the molarity and normality of the solution prepared by dissolving 55 mg of sodium bicarbonate (NaHCO3) in 750 mL of water is approximately 0.000873 mol/L and 0.000873 eq/L, respectively.